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THE NORMAL DISTRIBUTION Lesson 1

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Objectives To introduce the normal distribution The standard normal distribution Finding probabilities under the curve

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Normal Distributions Normal distributions are used to model continuous variables in many different situations. For example, a normal distribution could be used to model the height of students. We can transform our normal distribution into a standard normal distribution…

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The standard normal variable The curve is designed so that the total area underneath the curve is 1 The curve fits within ±4 standard deviations from the mean

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Find P (Z≤a) To do this we begin with a sketch of the normal distribution

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P (Z≤a) To do this we begin with a sketch of the normal distribution. We then mark a line to represent Z=a a

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P (Z≤a) To do this we begin with a sketch of the normal distribution. We then mark a line to represent Z=a P(Z≤a) is the area under the curve to the left of a. For continuous distributions there is no difference between P(Z≤a) and P(Z
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Ex1 Find P (Z<1.55) To do this we begin with a sketch of the normal distribution

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Ex 1 Find P (Z<1.55) To do this we begin with a sketch of the normal distribution. We then mark a line to represent Z=1.55 a

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Ex 1 Find P (Z<1.55) To do this we begin with a sketch of the normal distribution. We then mark a line to represent Z=a P(Z<1.55) is the area under the curve to the left of a. We now use the table to look up this probability a

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The Normal Distribution Table The table describes the positive half of the bell shaped curve… (Z) is sometimes used as shorthand for P(Z

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The Normal Distribution Table

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Ex 1 Find P (Z<1.55) To do this we begin with a sketch of the normal distribution. We then mark a line to represent Z=a P(Z<1.55) is the area under the curve to the left of a. We now use the table to look up this probability P(Z<1.55) = 0.9394 a

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Ex 2 Find P(Z>1.74)

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Note this result.. P(Z>a) = 1-P(Z1.74) = 1 – P(Z<1.74) = 1 – (0.9591) = 0.0409

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Ex 3 P(Z<-0.83) As our table only has values for the positive side of the distribution we must use symmetry…

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Ex 3 P(Z<-0.83) We have reflected the curve in the vertical axis. P(Z<-0.83) = 1- P(0.83) = 1 – (0.7967) = 0.2033 This is a really useful technique P(Z a) = 1- P(Z
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"description": "P(Z<-0.83) = 1- P(0.83) = 1 – (0.7967) = 0.2033 This is a really useful technique P(Z a) = 1- P(Z

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This is a really useful result P(Z<-a) = 1 - P(Z
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Ex 4 P(-1.24

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P(Z<2.16) = 0.9846

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Ex 4 P(-1.24

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Ex 4 P(-1.24

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Recommended work… Read through pages 177 and 178. Do Exercise 9A p179

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Blank Normal Distribution Plot

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